# surprisingly difficult filtration problem

I am interested in a proof of the following statement which seems intuitive, but is somehow really tricky:

Let $X$ be a stochastic process and let $(\mathcal{F}(t) : t \geq 0)$ be the filtration that it generates (unaugmented). Let $T$ be a bounded stopping time. Then we have $\mathcal{F}(T) = \sigma(X(T \wedge t) : t \geq 0)$

I have a proof at hand (Bain and Crisan, Fundamentals of Stochastic Filtering, page 309), but in my opinion there is a major gap. I will try to explain the idea of proof.

Let $V$ be the space of functions $[0,\infty) \rightarrow \mathbb{R}$ equipped with the sigma algebra generated by the cylinder sets. Consider the canonical map $X^T:\Omega \rightarrow V$ which maps $\omega$ to the trajectory $t \mapsto X(t \wedge T(\omega),\omega)$. Then we have $\sigma(X(T \wedge t) : t \geq 0) = \sigma(X^T)$.

The difficult part is $\subseteq$. Let $A \in \mathcal{F}(T)$. We want to find a measurable map $g:V \rightarrow \mathbb{R}$ such that $1_A = g \circ X^T$, then we're done. It is now straightforward to show that $1_A$ is constant on sets where the sample paths of $X^T$ are constant. (To be more precise: for $\rho \in \Omega$ consider the set $\mathcal{M}(\rho) = \lbrace \omega : X(\omega,t) = X(\rho,t), 0 \leq t \leq T(\rho) \rbrace$. Then $T$ and $1_A$ are constant on every set of this form).

The problem is: this is not sufficient! It suffices to construct a map $g$ such that $1_A = g \circ X^T$, but how we can we know that $g$ is measurable? This is where the proof of Bain and Crisan comes up short IMO.

I can show this result only under the assumption that the map $X:\Omega \rightarrow V$ be surjective: Since $A \in \mathcal{F}(\infty)$, we have a measurable map $g$ such that $1_A = g \circ X$. Let $\rho \in \Omega$. Then $T$ and $1_A$ are constant on $\mathcal{M}(\rho)$. Therefore, $g$ must be constant on the image of $\mathcal{M}(\rho)$ under $X$. Because $X$ is assumed to be surjective, this image contains the function $X^T(\rho)$. Hence, $g \circ X = g \circ X^T$, and we are done.

I think that this result could be a little bit deeper. I have seen two proofs of this for the special case that $X$ is the coordinate process on $C[0,\infty)$, one is given in the book of Karatzas & Shreve, Lemma 5.4.18. The fact that Karatzas proves this late in the book only in this special case somehow makes me think that the general case is not so easy.

I would really appreciate any comment or other reference for this result.

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There is an exercise in Kallenberg's Foundations of Modern Probability (exercise 6 of chapter 7, page 138 of the 2nd ed.) giving something in this direction. He assumes however that the process is progressive. The general result you mention would indeed be very nice but I also have doubts wether it holds in that generality. –  Jochen Wengenroth Jul 9 '12 at 7:18
As a matter of style, I don't see where you have used that the stopping time is bounded. I haven't had a chance to check but my guess is that Karatzas and Shreve treated the coordinate process because that ensures continuity of the process so you can approximate the process at discrete times which discretizes the stopping time and reduces the problem to one over a finite number of times. Every fix of your proof that I tried out last night needed some sort of regularity for the process so I don't have anything else to offer you. –  BSteinhurst Jul 9 '12 at 12:54
After havin a look at the book, I do not trust the proof in Bain and Crisan either. On a formal level it seems like a trivial fault: They deduce necessary conditions in A.22. and A.23. which they verify in the sufficiency part of the proof -- of course, this is not conclusive. The way you describe the argument shows the difficulty if one does not have any regularity: Measurability of the function $g: \mathbb{R}^{[0,\infty)} \to \mathbb{R}$ is a very restrictive condition because $g$ must only depend on countably many variables. –  Jochen Wengenroth Jul 9 '12 at 14:38
I don't think this proof can be fixed, which is really unsatisfactory, because this result is crucial in deriving the filtering equations using the innovations approach. I will try to prove it in the continuous case. But even there, I don't think this direct approach will work. –  Hauke L. Jul 9 '12 at 18:16
In fact, Bain and Crisan don't even use that the stopping time is bounded. What they claim is that, for a stopping time $T$, we have $\mathcal{F}(t \wedge T) = \sigma ( X(s \wedge T) : 0 \leq s \leq t)$, i.e., the "stopped filtration" equals the filtration generated by the stopped process. This result follows directly if the above one is true. They could assume in the proof that $T$ is bounded, but they don't do that. –  Hauke L. Jul 9 '12 at 18:25